a block sits at rest on a horizontal surface. one end of a spring is attached to the block and the other end of the spring is attached to a wall, as shown in figure 1. the block is initially at position ( x = 0 ), where the spring is at its natural length. the block is then pushed to the left, compressing the spring. the block is released from rest from position ( x = -a ). all frictional forces are negligible.
consider the system that consists of the block and the spring. figure 2 shows an energy bar chart that represents the kinetic energy ( k ) and the elastic potential energy ( u_s ) of the system when the block is released from rest at position ( x = -a ).
(a) draw shaded bars that represent ( k ) and ( u_s ) of the system to complete the energy bar charts shown in figure 3 for when the block is at positions ( x = 0 ) and ( x = +a/2 ).
- shaded regions should start at the solid line that represents zero energy.
- represent any energy that is equal to zero with a distinct line on the zero - energy line.
- the relative heights of each shaded region should reflect the magnitudes of the respective energy consistent with the scale used in figure 2.
note on your ap exam, you will handwrite your responses to free - response questions in a test booklet
(b) the mass of the block is ( m ), and the spring constant of the spring is ( k ). starting with conservation of energy, derive an expression for the speed of the block as it passes through position ( x = +a/2 ). express your answer in terms of ( m ), ( k ), ( a ) and physical constants, as appropriate.

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Accepted Answer

# Explanation:
## Step1: Write the conservation of energy equation
The total mechanical energy $E$ of the system is conserved. At $x = - A$, the kinetic energy $K_{i}=0$ and the elastic potential energy $U_{si}=\frac{1}{2}kA^{2}$. At $x=\frac{A}{2}$, the kinetic energy $K_{f}=\frac{1}{2}mv^{2}$ and the elastic potential energy $U_{sf}=\frac{1}{2}k(\frac{A}{2})^{2}$. According to the conservation of energy $E = K_{i}+U_{si}=K_{f}+U_{sf}$, so $0+\frac{1}{2}kA^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}k(\frac{A}{2})^{2}$.
## Step2: Solve the equation for $v$
First, simplify the energy - conservation equation:
$$
\begin{align*}
\frac{1}{2}kA^{2}&=\frac{1}{2}mv^{2}+\frac{1}{2}k\frac{A^{2}}{4}\
\frac{1}{2}kA^{2}-\frac{1}{8}kA^{2}&=\frac{1}{2}mv^{2}\
\frac{4kA^{2}-kA^{2}}{8}&=\frac{1}{2}mv^{2}\
\frac{3kA^{2}}{8}&=\frac{1}{2}mv^{2}\
v^{2}&=\frac{3kA^{2}}{4m}\
v&=\frac{A}{2}\sqrt{\frac{3k}{m}}
\end{align*}
$$

# Answer:
$v = \frac{A}{2}\sqrt{\frac{3k}{m}}$

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Explanation:

Step1: Write the conservation of energy equation

Step2: Solve the equation for \(v\)

Answer: